Timeline for On closed totally disconnected subgroups of connected real Lie groups
Current License: CC BY-SA 3.0
14 events
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| Apr 16, 2011 at 18:34 | comment | added | agt | @Hugo Chapdelaine: does not your question concern Lie groups? | |
| Apr 16, 2011 at 18:31 | history | edited | agt | CC BY-SA 3.0 |
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| Apr 16, 2011 at 18:30 | comment | added | Hugo Chapdelaine | well you just added in your statement the assumption that $G$ is locally euclidean, so now it is fine! | |
| Apr 16, 2011 at 18:08 | history | edited | agt | CC BY-SA 3.0 |
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| Apr 16, 2011 at 17:53 | comment | added | agt | @Hugo Capdelaine but in such a case even $G$ would be totally disconnected, while you assume $G$ connected | |
| Apr 16, 2011 at 17:43 | comment | added | agt | @Hugo Chapdelaine: as outlined in my answer, when $H$ is a closed subgroup of the Lie Group $G$, the Cartan--Von Neumann theorem implies that $H$ is an embedded Lie group. Aside it is clear, from the definition, that a topological manifold is totally disconnected if and only if it is 0-dimensional. | |
| Apr 16, 2011 at 17:35 | comment | added | Hugo Chapdelaine | Well take $G=H=\mathbf{Z}_p$, then $H$ is not discrete. The result that you claim in $0)$ probably applies to topological groups which have a topological real manifold structure. | |
| Apr 16, 2011 at 17:33 | comment | added | Qiaochu Yuan | @Hugo: exactly what it sounds like: a (second-countable Hausdorff etc.) space which is locally homeomorphic to a point, hence discrete. | |
| Apr 16, 2011 at 17:25 | history | edited | agt | CC BY-SA 3.0 |
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| Apr 16, 2011 at 16:55 | comment | added | Hugo Chapdelaine | So what is the definition of a 0-dimensional manifold? | |
| Apr 16, 2011 at 16:42 | comment | added | Brad Hannigan-Daley | Cartan's theorem implies that $H$ is necessarily a smooth submanifold of $G$. So if it's totally disconnected, it must be 0-dimensional, i.e. discrete. | |
| Apr 16, 2011 at 16:22 | comment | added | Hugo Chapdelaine | Hi Giuseppe, but $H$ is not necessarily a topological manifold. So are you saying that $H$ is a $0$-dimensional manifold so therefore discrete? | |
| Apr 16, 2011 at 15:50 | history | edited | agt | CC BY-SA 3.0 |
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| Apr 16, 2011 at 15:42 | history | answered | agt | CC BY-SA 3.0 |